Question

A small coin, initially at rest, begins falling. If the clock starts when the coin begins to fall, what is the magnitude of the coin\'s displacement between t1 = 0.2212 s and t2 = 0.737 s?

Answer 1

As per question the coin is falling under gravity as there is no other force that has bee mentioned here.

so the equation of motion for a freely falling body is given as-$s=ut-\frac{1}{2} gt^2$

here the initial velocity is zero as the coin was at rest.

hence we have $s=\frac{1}{2} gt^2$

so now we have to calculate the displacement of the coin at various instants.

at t=0.2212 s,we have

     $s=\frac{1}{2} 9.801*[0.2212]^2$

              =0.2398m[upto four decimal digit

again when t=0.737 s

s=$\frac{1}{2} *9.801*[0.737]^2$

 =2.662m [upto three decimal digit]  [ans]